1. **Rewrite without the absolute value sign:** (a) Given $|x| < 3$, this means $-3 < x < 3$. (b) Given $|x| - 31 < 5$, add 31 to both sides: $|x| < 36$, so $-36 < x < 36$. (c) Given $|2x - 5| < 9$, this means $-9 < 2x - 5 < 9$. Add 5 to all parts: $-4 < 2x < 14$. Divide by 2: $-2 < x < 7$. 2. **Find integers $q$ and $r$ such that $a = bq + r$ and $0 \leq r < |b|$:** (a) $a=258$, $b=12$. Divide 258 by 12: $258 \div 12 = 21$ remainder $6$. So $q=21$, $r=6$. (b) $a=395$, $b=14$. Divide 395 by 14: $395 \div 14 = 28$ remainder $3$. So $q=28$, $r=3$. (c) $a=608$, $b=-17$. Divide 608 by 17: $608 \div 17 = 35$ remainder $13$ (since $17 \times 35 = 595$, $608-595=13$). Since $b$ is negative, $|b|=17$, and $0 \leq r < 17$ holds. So $q=35$, $r=13$. 3. **Write each open interval in the form $|x - a| < r$:** (a) $3 < x < 9$. Midpoint $a = \frac{3+9}{2} = 6$, radius $r = 9 - 6 = 3$. So $|x - 6| < 3$. (b) $-5 < x < 1$. Midpoint $a = \frac{-5+1}{2} = -2$, radius $r = 1 - (-2) = 3$. So $|x + 2| < 3$. 4. **Find gcd, integers $m,n$ such that $d=ma+nb$, and lcm:** Given $a=5880$, $b=8316$. (a) Find $d = \gcd(5880, 8316)$ using Euclidean algorithm: $8316 = 5880 \times 1 + 2436$ $5880 = 2436 \times 2 + 1008$ $2436 = 1008 \times 2 + 420$ $1008 = 420 \times 2 + 168$ $420 = 168 \times 2 + 84$ $168 = 84 \times 2 + 0$ So $d = 84$. (b) Find $m,n$ such that $84 = m \times 5880 + n \times 8316$ by back substitution: $84 = 420 - 168 \times 2$ $168 = 1008 - 420 \times 2$ $420 = 2436 - 1008 \times 2$ $1008 = 5880 - 2436 \times 2$ $2436 = 8316 - 5880$ Substitute stepwise: $84 = 420 - 2(1008 - 420 \times 2) = 5 \times 420 - 2 \times 1008$ $= 5(2436 - 2 \times 1008) - 2 \times 1008 = 5 \times 2436 - 12 \times 1008$ $= 5 \times 2436 - 12(5880 - 2 \times 2436) = 29 \times 2436 - 12 \times 5880$ $= 29(8316 - 5880) - 12 \times 5880 = 29 \times 8316 - 41 \times 5880$ So $m = -41$, $n = 29$. (c) Find $\mathrm{lcm}(a,b)$ using $\mathrm{lcm}(a,b) = \frac{|a b|}{\gcd(a,b)}$: $\mathrm{lcm}(5880, 8316) = \frac{5880 \times 8316}{84} = 582120$. 5. **Clara's pizza and cola problem:** Let $p$ = number of pizzas, $c$ = number of cola bottles. Each pizza costs 57, each cola costs 22, total money = 400. Inequality: $57p + 22c \leq 400$. To maximize purchases, try integer values: For $p=0$, max $c = \lfloor 400/22 \rfloor = 18$. For $p=1$, $22c \leq 400 - 57 = 343$, $c = \lfloor 343/22 \rfloor = 15$. For $p=2$, $22c \leq 400 - 114 = 286$, $c = 13$. For $p=3$, $22c \leq 400 - 171 = 229$, $c = 10$. For $p=4$, $22c \leq 400 - 228 = 172$, $c = 7$. For $p=5$, $22c \leq 400 - 285 = 115$, $c = 5$. For $p=6$, $22c \leq 400 - 342 = 58$, $c = 2$. For $p=7$, $22c \leq 400 - 399 = 1$, $c = 0$. So Clara can buy combinations like $(p,c) = (7,0), (6,2), (5,5), (4,7), (3,10), (2,13), (1,15), (0,18)$. 6. **Prove $(A \times B) \cap (A \times C) = A \times (B \cap C)$:** To prove set equality, show both inclusions: - If $(a,x) \in (A \times B) \cap (A \times C)$, then $(a,x) \in A \times B$ and $(a,x) \in A \times C$. This means $a \in A$, $x \in B$, and $x \in C$, so $x \in B \cap C$. Hence $(a,x) \in A \times (B \cap C)$. - Conversely, if $(a,x) \in A \times (B \cap C)$, then $a \in A$ and $x \in B \cap C$, so $x \in B$ and $x \in C$. Thus $(a,x) \in A \times B$ and $(a,x) \in A \times C$, so $(a,x) \in (A \times B) \cap (A \times C)$. Therefore, $(A \times B) \cap (A \times C) = A \times (B \cap C)$. **Final answers:** (a) $|x| < 3 \Rightarrow -3 < x < 3$ (b) $|x| - 31 < 5 \Rightarrow -36 < x < 36$ (c) $|2x - 5| < 9 \Rightarrow -2 < x < 7$ 2(a) $q=21$, $r=6$ 2(b) $q=28$, $r=3$ 2(c) $q=35$, $r=13$ 3(a) $|x - 6| < 3$ 3(b) $|x + 2| < 3$ 4(a) $d=84$ 4(b) $m=-41$, $n=29$ 4(c) $\mathrm{lcm} = 582120$ 5) Possible $(p,c)$ pairs: $(7,0), (6,2), (5,5), (4,7), (3,10), (2,13), (1,15), (0,18)$ 6) $(A \times B) \cap (A \times C) = A \times (B \cap C)$ proven.